Why is the Big-O complexity of this algorithm O(n^2)?

Question

I know the big-O complexity of this algorithm is O(n^2), but I cannot understand why.

int sum = 0; 
int i = 1; j = n * n; 
while (i++ < j--) 
  s         


        

Answer 1

Let m be the number of iterations taken. Then,

i+m = n^2 - m

which gives,

m = (n^2-i)/2

In Big-O notation, this implies a complexity of O(n^2).